Sulfur trioxide, SO3 , is produced in enormous quantities each year for use in the synthesis of sulfuric acid. S(s) + O2(g) > SO2(g)
2SO2(g) + O2(g) > 2SO3(g)
What volume of O2(g) at 340.°С and a pressure of 4.75 atm is needed to completely convert 6.30 g sulfur to sulfur trioxide?
Volume = L

3.14 L of oxygen (O₂).

Explanation:

We'll begin by calculating the number of mole in 6.3 g of sulphur (S). This can be obtained as follow:

Molar mass of S = 32 g/mol

Mass of S = 6.3 g

Mole of S =.?

Mole = mass / molar mass

Mole of S = 6.3/32

Mole of S = 0.197 mole

Next, we shall write the overall equation of the reaction between sulphur (S) and oxygen (O₂) to produce sulphur trioxide (SO₃) .

This is illustrated below:

S (s) + O₂ (g) —> SO₂ (g)

SO₂ (g) + O₂ (g) —> 2SO₃ (s)

Overall reaction:

2S (s) + 3O₂ (g) —> 2SO₃ (g)

Next, we shall determine the number of mole of oxygen (O₂) needed to completely convert 6.30 g (i.e 0.197 mole) of sulfur.

This is illustrated below:

From the balanced equation above,

2 moles of sulphur (S) required 3 moles of oxygen (O₂) .

Therefore, 0.197 mole of sulphur (S) will require = (0.197 × 3)/2 = 0.296 mole of oxygen (O₂).

Therefore, 0.296 mole of oxygen (O₂) is needed.

Finally, we shall determine the volume of oxygen (O₂) needed as follow:

Number of mole (n) of oxygen (O₂) = 0.296 mole

Temperature (T) = 340 °С = 340 °С + 273 = 613 K

Pressure (P) = 4.75 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) of oxygen (O₂) =.?

PV = nRT

4.75 × V = 0.296 × 0.0821 × 613

Divide both side by 4.75

V = (0.296 × 0.0821 × 613) / 4.75

V = 3.14 L

Therefore, 3.14 L of oxygen (O₂) is needed for the reaction.

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole